Since the genus of \(\Sigma \)  is \(\mathrm {g}\geqslant 2\) , we can find a constant curvature \(-1\)  metric \(g_\ell \)  on \(\Sigma \) , such that the closed geodesic \(\gamma \in h\)  has length \(\ell >0\) . By the Collar Lemma [1]}, provided \(\ell >0\)  is small enough, \(\gamma \)  has a neighbourhood of the form \((\mathbb {R}/\ell \mathbb {Z})_\tau \times [-2, 2]_\rho \)  where the metric writes
\(\mathrm {d}s^2 = h(\rho )^2\mathrm {d}\tau ^2 + \mathrm {d}\rho ^2,\) 
where
\(h(\rho ) = \cosh \rho .\) 
Let \(H = (R_{-\pi / 2})_* X\)  and \(V\)  be the generator of the flow \((R_\theta )_{\theta \in \mathbb {R}}\) .
In the coordinates \((\partial _\tau , \partial _\rho , \partial _\theta )\) , we have \(V = \partial _\theta \) ,
\(X =\begin{pmatrix} h^{-1}\cos \theta \\ \sin \theta \\ 4 \cos \theta \, \tanh \rho \end{pmatrix} \quad \text{ and } \quad H =\begin{pmatrix} -h^{-1}\sin \theta \\ \cos \theta \\ -4 \sin \theta \, \tanh \rho \end{pmatrix}.\) 
Moreover, because \(\kappa = -1\) , it is a classical result that
\(E_s = \mathbb {R}(H - V) \quad \text{and} \quad E_u = \mathbb {R}(H + V).\) 
On the other hand, let \(\chi \in C^\infty ([-2, 2], [0, 1])\)  be a monotone function such that \(\chi \equiv 0\)  on \([-2, -1]\)  and \(\chi \equiv 1\)  on \([1, 2]\) , and put
\(f_0(\tau , \rho ) = (\tau + \ell \chi (\rho ), \rho ).\) 
Now set
\(F_0(\tau , \rho , v) = \left(f_0(\tau , \rho ), \frac{\mathrm {d}f_0(\tau , \rho ) v}{\Vert \mathrm {d}f_0(\tau , \rho ) v\Vert }\right).\) 
Then it is not hard to see that, in the coordinates \((\tau , \rho , \theta )\) , we have
\(F_0(\tau , \rho , \theta ) = \left(\tau + \ell \chi (\rho ), \rho , \varphi \right)\) 
where \(\varphi \in \mathbb {R}/2\pi \mathbb {Z}\)  is defined by
\(\tan \varphi = \frac{\sin \theta }{\cos \theta + \ell \chi ^{\prime }(\rho )h(\rho )\sin \theta }\) 
if \(\chi ^{\prime }(\rho ) \ne 0\)  and \(\varphi = \theta \)  otherwise.
We may thus write
\(\mathrm {d}F_0(\tau , \rho , \theta ) =\begin{pmatrix}1 & \ell \chi ^{\prime }(\rho ) & 0 \\ 0 & 1 & 0 \\ 0 & \partial _\rho \varphi & \partial _\theta \varphi \end{pmatrix}.\) 
For \(z = (\tau , \rho , \theta )\)  let \(\beta _z = (\partial _\tau , \partial _\rho , \partial _\theta )\) , and \(\tilde{\beta }_z = (X(z), H(z), K(z))\) . Then the matrix of \(\tilde{\beta }_z\)  in \(\beta _z\)  is given by
\([\beta _z : \tilde{\beta }_z] =\begin{pmatrix}h^{-1} \cos \theta & -h^{-1} \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 2 \cos \theta \, \tanh \rho & -2 \sin \theta \, \tanh \rho & 1\end{pmatrix},\) 
and its inverse \([\tilde{\beta }_{F_0(z)} : \beta _{F_0(z)}] \)  at \(F_0(z) = \left(\tau + \ell \chi (\rho ), \rho , \varphi \right)\)  reads
\(\begin{pmatrix}h \cos \varphi & \sin \varphi & 0 \\ -h \sin \varphi & \cos \varphi & 0 \\ -2 \partial _\rho h & 0 & 1\end{pmatrix}.\) 
Using (REF ) we can compute the matrix of \(\mathrm {d}F_0(z)\)  computed in the basis \(\tilde{\beta }_z\)  and \(\tilde{\beta }_{F_0(z)}\) ; it is of the form
\(\begin{pmatrix}\lambda & \mu ^\perp & 0 \\0 & \mu & 0 \\\delta & \varepsilon & \nu \end{pmatrix}\) 
for some functions \(\lambda , \mu ^\perp , \mu , \delta , \varepsilon , \nu \)  of \((\tau , \rho , \theta )\) , with \(\nu = \partial _\theta \varphi \)  and
\(\begin{aligned}\mu &= \sin \theta \sin \varphi + \cos \theta (-\ell h \chi ^{\prime } \sin \varphi + \cos \varphi ), \\\varepsilon &= -2 \partial _\rho h(-h^{-1} \sin \theta + \ell \chi ^{\prime } \cos \theta ) - 2 h^{-1} \partial _\rho h \sin \theta \partial _\theta \varphi + \partial _\rho \varphi \cos \theta .\end{aligned}\) 
Now, write
\(\varepsilon = 2 h^{-1} \partial _h \sin \theta (1 - \partial _\theta \varphi ) + \partial _\rho \varphi \cos \theta - 2 \ell \chi ^{\prime } \partial _\rho h \cos \theta .\) 
Equation (REF ) yields
\(\partial _\theta \varphi \rightarrow 1, \quad \partial _\rho \varphi \rightarrow 0 \quad \text{and} \quad \varphi \rightarrow \theta \) 
uniformly on \([-2, 2] \times \mathbb {R}/\ell \mathbb {Z}\)  as \(\ell \rightarrow 0\) . Therefore we get
\(\varepsilon \rightarrow 0 \quad \text{and} \quad \mu , \nu \rightarrow 1\) 
as \(\ell \rightarrow 0.\)  Consequently, whenever \(\ell \)  is small enough, we have
\(\mathrm {d}F_0(E_u \oplus E_0) \ne E_s \oplus E_0\) 
by (REF ), which is equivalent to the condition  REF. Since  REF is an open condition, any sufficiently small perturbation of \(F_0\)  will still satisfy it.
We can find \(f\)  arbitrarily close to \(f_0\) , with non degenerate fixed points \(x_1, \dots , x_q\) . Let \(F_1\)  be a bundle map over \(f\) , close to \(F_0\) , so that \(F_1\)  satisfies  REF.
Next, consider the function
\(F : \mathbb {R}/2\pi \mathbb {Z}\times M \rightarrow M \times M\) 
defined by
\(F(\theta , z) = (R_\theta F_1(z), z),\) 
and set \(F_\theta (z) = R_\theta \circ F_1\) . We claim that for a dense set of \(\theta \) 's, \(F_\theta \)  satisfies  REF and  REF. For this we will use the tranversality result [2]}.
For point  REF, observe that \(F_\theta \)  has non degenerate fixed points if and only if \(F_\theta \)  has non-degenerate fixed points in each fiber \(S_{x_j} \Sigma \) , \(j=1\dots q\) . However,
\(\mathrm {ran} \left(\mathrm {d}{F}_{|\mathbb {R}\times S_{x_j}\Sigma }\right) = T [S_{x_j}\Sigma ] ^2\) 
is tranverse to the diagonal, so that according to [2]}, for almost every \(\theta \) , \(F_\theta \)  has non degenerate fixed points in \(S_{x_j} \Sigma \) .
Let us turn to point  REF. Define for \(\tau >0\)
\(G_\tau = \lbrace (\varphi _t(z), z)\ |\ 0 \leqslant t < \tau ,~z \in M\rbrace  \subset M \times M.\) 
Let us prove that \({F}\)  is transverse to \(G_\tau \) . Let \(z \in M\) , \(\theta \in \mathbb {R}/2\pi \mathbb {Z}\)  and \(\tau > t\geqslant 0\)  such that we have \(R_\theta F_1(x,v) = \varphi _t(z)\) . For simplicity we set \(z_t = \varphi _t(z)\) . If \(t=0\) , the tranversality of \({F}\)  with \(G_\varphi \)  follows from the tranversality with the diagonal, so we assume that \(t>0\) . At our point of intersection, \(\mathrm {ran} (\mathrm {d}{F})\)  contains
\(T_{z_t} S_{f(x)} \Sigma \times T_{z} S_x \Sigma = \mathrm {vect} \lbrace  (V,0),\ (0,V) \rbrace \) 
On the other and, \(T G_\tau \)  is generated by \((X, 0)\)  and the vectors
\((\mathrm {d}\varphi _t(z)W, W), \quad W \in T_{z} M.\) 
In particular, \(T G_\tau \)  contains also \((0,X)\) . Finally, since \(\Sigma \)  has no conjugate points, \(\mathrm {d}\varphi _t(V)\)  always has a non-zero component along \(H\)  when \(t>0\) , so that \(\mathrm {ran} (\mathrm {d}{F}) + T G_\tau \)  contains
\((X,0),\ (0,X),\ (V,0),\ (0,V),\ (H,0),\ (0,H),\) 
which generate \(T_{(z_t, z)}(M \times M)\) . Taking a sequence \(\tau =n\) , \(n\in \mathbb {N}\) , we can use again [2]} to conclude. This closes the proof of Proposition REF .
